AQA Paper 2 2018 June — Question 16 6 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjection from elevated point - angle above horizontal
DifficultyStandard +0.3 This is a standard projectile motion problem requiring straightforward application of kinematic equations. Part (a) uses the fact that vertical velocity is zero at the highest point to find u from v = u + at. Part (b) requires solving a quadratic equation for time of flight using s = ut + ½at². While it involves multiple steps and careful sign handling, it follows a routine textbook approach with no novel insight required, making it slightly easier than average.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model
In this question use \(g = 9.81\) m s\(^{-2}\) A particle is projected with an initial speed \(u\), at an angle of 35° above the horizontal. It lands at a point 10 metres vertically below its starting position. The particle takes 1.5 seconds to reach the highest point of its trajectory.
  1. Find \(u\). [3 marks]
  2. Find the total time that the particle is in flight. [3 marks]
Question 16:
AnswerMarks
16(a)Uses vuatwith v0 for
the vertical motion
AnswerMarks Guidance
Condone cos or sign errorAO3.4 M1
(cid:1873) (cid:3404) 25.7 m s-1
AnswerMarks Guidance
Obtains correct equationAO1.1b A1
Obtains correct (cid:1873) to 3AO1.1b A1
significant figures
CAO
AnswerMarks
16(b)1
Uses s ut at2
2
with s10 and their ufor
vertical motion
AnswerMarks Guidance
Condone cos or sign errorAO3.4 M1
1025.7sin35t 9.81t2
2
t = 3.571
Time in flight is 3.57 seconds
AnswerMarks Guidance
Obtains correct equationAO1.1b A1F
Obtains correct time of flight
with units
AWRT 3.6
AnswerMarks Guidance
CAOAO3.2a A1
Total6
AO1.1b
A1
AnswerMarks Guidance
QMarking Instructions AO 
Question 16:
--- 16(a) ---
16(a) | Uses vuatwith v0 for
the vertical motion
Condone cos or sign error | AO3.4 | M1 | 0 = u sin 35 – 9.81 x 1.5
(cid:1873) (cid:3404) 25.7 m s-1
Obtains correct equation | AO1.1b | A1
Obtains correct (cid:1873) to 3 | AO1.1b | A1
significant figures
CAO
--- 16(b) ---
16(b) | 1
Uses s ut at2
2
with s10 and their ufor
vertical motion
Condone cos or sign error | AO3.4 | M1 | 1
1025.7sin35t 9.81t2
2
t = 3.571
Time in flight is 3.57 seconds
Obtains correct equation | AO1.1b | A1F
Obtains correct time of flight
with units
AWRT 3.6
CAO | AO3.2a | A1
Total | 6
AO1.1b
A1
Q | Marking Instructions | AO | Marks | Typical Solution
In this question use $g = 9.81$ m s$^{-2}$

A particle is projected with an initial speed $u$, at an angle of 35° above the horizontal.

It lands at a point 10 metres vertically below its starting position.

The particle takes 1.5 seconds to reach the highest point of its trajectory.

\begin{enumerate}[label=(\alph*)]
\item Find $u$.
[3 marks]

\item Find the total time that the particle is in flight.
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2 2018 Q16 [6]}}
This paper (17 questions)
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Q1 1 Q2 1 Q3 1 Q4 6 Q5 2 Q6 7 Q7 8 Q8 10 Q9 14 Q10 1 Q11 1 Q12 5 Q13 8 Q14 6 Q15 9 Q16 6 Q17 14